If iodine has a 13.3 hour half life, 3.12 mg of the original sample will still be present after 39.9 hours.
<h3>what is The meaning of half-life?</h3>
The reaction rate must decrease to half its initial value before the process has reached its half life. The split of that first order chemical reaction is a quantity related to the rate constant of the reaction.
Briefing:
Given, the iodine half-life is 13.3 hours. As a result, half-life t1/2 Equals ln 2/ K
k = 0.693 .... t1/2
≈ 0.693 /13.3 hours
= 5.21 *10-2
The rate law sets the flow timetime is set by the rate law, which varies for each emotion. When only one reactant's quantity influences the speed of chemical reaction, this is referred to as a first-order contact. Because of this, it is occasionally referred to as unimolecular process. The rate is as follows for this type of reaction: k= 1/t ln [A]0 / [A] is the rate constant for the first order reaction. 5.21 * 10-2 hours are in 39.9 hours.
By including all the numbers in the initial order reaction's rate constant equation, we obtain [A].
t = 3.12mg
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the answer is true the first one
It has an ultramafic composition
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Answer:
3.568 g of H₂O
Solution:
The Balance Chemical Equation is as follow,
2 H₂ + O₂ → 2 H₂O
Step 1: Calculate the Limiting Reagent,
According to Balance equation,
4.04 g (2 mol) H₂ reacts with = 32 g (1 mol) of O₂
So,
0.40 g of H₂ will react with = X g of O₂
Solving for X,
X = (0.40 g × 32 g) ÷ 4.04 g
X = 3.17 g of O₂
It means 0.4 g of H₂ requires 3.17 g of O₂, while we are provided with 3.2 g of O₂ which is in excess. Therefore, H₂ is the limiting reagent and will control the yield of products.
Step 2: Calculate amount of Water produced,
According to equation,
4.04 g (2 mol) of H₂ produces = 36.04 g (2 mol) of H₂O
So,
0.40 g of H₂ will produce = X moles of H₂O
Solving for X,
X = (0.40 g × 36.04 mol) ÷ 4.04 g
X = 3.568 g of H₂O
<span>As mentioned, the isomerization of cyclopropane to propylene is a first-order process with a half-life of 19 min at 500°c. A first-order reaction kinetic rates means that the rate is constant throughout the reaction.
Thus, the time it takes for the partial pressure of cyclopropane to decrease from 1 atm to 0.125 atm at 500°c is </span><span>57 minutes.</span>